Question: Find the distance between the point ${(1, 1)}$ and the line $\enspace {y = \dfrac{1}{3}x + 4}\thinspace$. {1} {2} {3} {4} {5} {6} {7} {8} {9} {\llap{-}2} {\llap{-}3} {\llap{-}4} {\llap{-}5} {\llap{-}6} {\llap{-}7} {\llap{-}8} {\llap{-}9} {1} {2} {3} {4} {5} {6} {7} {8} {9} {\llap{-}2} {\llap{-}3} {\llap{-}4} {\llap{-}5} {\llap{-}6} {\llap{-}7} {\llap{-}8} {\llap{-}9}
Answer: First, find the equation of the perpendicular line that passes through ${(1, 1)}$ The slope of the blue line is ${\dfrac{1}{3}}$ , and its negative reciprocal is ${-3}$ Thus, the equation of our perpendicular line will be of the form $\enspace {y = -3x + b}\thinspace$ We can plug our point, ${(1, 1)}$ , into this equation to solve for ${b}$ , the y-intercept. $1 = {-3}(1) + {b}$ $1 = -3 + {b}$ $1 + 3 = {b} = 4$ The equation of the perpendicular line is $\enspace {y = -3x + 4}\thinspace$ We can see from the graph (or by setting the equations equal to one another) that the two lines intersect at the point ${(0, 4)}$ . Thus, the distance we're looking for is the distance between the two red points. The distance formula tells us that the distance between two points is equal to: $\sqrt{( x_{1} - x_{2} )^2 + ( y_{1} - y_{2} )^2}$ Plugging in our points ${(1, 1)}$ and ${(0, 4)}$ gives us: $\sqrt{( {1} - {0} )^2 + ( {1} - {4} )^2}$ $= \sqrt{( 1 )^2 + ( -3 )^2} = \sqrt{10} $ The distance between the point ${(1, 1)}$ and the line $\thinspace {y = \dfrac{1}{3}x + 4}\enspace$ is $\thinspace\sqrt{10}$.